NEET physics

Question 1

If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are

  • [FvT-1]

  • [FvT-2]

  • [Fv-1T-1]

  • [Fv-1T]

Solution

D.

[Fv-1T]

We know that,
F = ma
That is,
F = mv/t
straight m space equals space Ft over straight v
So, left square bracket straight M right square bracket space equals space fraction numerator left square bracket straight F right square bracket left square bracket straight T right square bracket over denominator left square bracket straight V right square bracket end fraction space equals space left square bracket space Fv to the power of negative 1 end exponent straight T right square bracket

Question 2

A projectile is fired from the surface of the earth with a velocity of 5 m/s and angle straight theta with the horizontal. Another projectile fired from another planet with a velocity of 3 m/s at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is,

  • 3.5

  • 5.9

  • 16.3

  • 110.8

Solution

A.

3.5

If the trajectory is same for both the projectiles, their maximum height will be the same.
That is,
(Hmax)1 = (Hmax)2
i.e., 
space space space space space space space space fraction numerator straight u subscript 1 squared space sin squared straight theta over denominator 2 straight g subscript 1 end fraction space equals space fraction numerator straight u subscript 2 squared space sin squared straight theta over denominator 2 straight g subscript 2 end fraction
rightwards double arrow space straight u subscript 1 squared over straight u subscript 2 squared space equals space straight g subscript 1 over straight g subscript 2
rightwards double arrow space fraction numerator left parenthesis 5 right parenthesis squared over denominator left parenthesis 3 right parenthesis squared end fraction space equals space fraction numerator 9.8 over denominator g subscript 2 end fraction
rightwards double arrow space g subscript 2 italic space italic equals italic space fraction numerator italic 9 italic. italic 8 italic space x italic space italic 9 over denominator italic 25 end fraction
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic equals italic space italic 3 italic. italic 5 italic space m italic divided by s to the power of italic 2

Question 3

A particle is moving such that its position coordinates (x,y) are (2m, 3m) at time t = 0, (6m, 7m) at time t = 2 sec and (13 m, 14m) at time t= 5 sec. Average velocity vector (vav) from t = 0 to t = 5 s is,

  • 1 fifth left parenthesis 13 space i with hat on top space plus space 14 space j with hat on top right parenthesis
  • 7 over 3 open parentheses i with rightwards arrow on top space plus space j with hat on top close parentheses
  • 2 space left parenthesis straight i with hat on top space plus space straight j with hat on top right parenthesis
  • 11 over 5 open parentheses i with hat on top plus j with hat on top close parentheses

Solution

D.

11 over 5 open parentheses i with hat on top plus j with hat on top close parentheses

Average velocity vector,

straight v subscript av space equals space fraction numerator Net space displacement over denominator Time space taken end fraction

space space space space space space equals space fraction numerator left parenthesis 13 minus 2 right parenthesis straight i with hat on top space plus space left parenthesis 14 minus 3 right parenthesis straight j with hat on top over denominator 5 end fraction

space space space space space space equals space fraction numerator 11 space straight i with hat on top space plus space 11 space straight j with hat on top over denominator 5 end fraction

space space space space space space space equals space 11 over 5 open parentheses straight i with hat on top space plus space straight j with hat on top close parentheses

Question 4

A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction= straight mu). the pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is,

  • fraction numerator straight g space left parenthesis 1 minus gμ right parenthesis over denominator straight g end fraction
  • fraction numerator 2 gμ over denominator 3 end fraction
  • fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction
  • fraction numerator straight g left parenthesis 1 minus 2 straight mu right parenthesis over denominator 2 end fraction

Solution

C.

fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction

First of all consider the forces on the blocks,

For the 1st block,
mg - T1 = m x a        ... (i)
Let us consider second and third block as a system,
Then,
T12 μmg = 2m x a
mg (1 space minus 2 straight mu) = 3m x a
straight a space equals space 2 over 3 left parenthesis 1 minus 2 straight mu right parenthesis 

Question 5

The force F acting on a particle of mass m is indicated by the force-time graph as shown below. The climate change in momentum of the particle over the time interval from zero to 8 s is,

  • 24 Ns

  • 20 Ns

  • 12 Ns

  • 6 Ns

Solution

C.

12 Ns

The area under F-t graph gives change in momentum.
For 0 to 2 s,
increment straight p subscript 1 space equals space 1 half straight x 2 straight x 6 space equals space 6 space kg minus straight m divided by straight s
For 2 to 4 sec,
increment straight p subscript 2 space equals space 2 space straight x space left parenthesis negative 3 right parenthesis space equals space minus 6 space kg minus straight m divided by straight s
For 4 to 8 sec,
increment straight p subscript 3 space equals space left parenthesis 4 right parenthesis straight x 3 space equals space plus 12 space kg minus straight m divided by straight s
So, total change in momentum for 0 to 8 sec is,
increment straight p subscript net space equals space increment straight p subscript 1 plus increment straight p subscript 2 plus increment straight p subscript 3

space space space space space space space space space space space equals left parenthesis plus 6 space minus 6 space plus 12 right parenthesis

space space space space space space space space space space space equals space 12 space kg minus straight m divided by straight s

space space space space space space space space space space space equals space 12 space Ns